Why the lens space L(2,1) is homeomorphic to $\mathbb{R}P^3$?

Question

According to one definition of lens space $L(p,q)$, which is gluing two solid tori with a map $h:T^2_1 \rightarrow T^2_2$. And $h(m_1)=pl_2+qm_2$, $l_i$ means longitude and $m_i$ means meridian of the boundary torus. I cannot understand why $L(2,1)$ is homeomorphic to $\mathbb{R}P^3$. Can someone give me some hints? Thank you.

Answer 1

Here's a sort of diagrammatic argument from an old homework assignment:

Construct $\mathbb{R}P^3$ as the quotient of $B^3 \subset \mathbb{R}^3$ under the antipodal map $a: \partial B^3 \to \partial B^3$. Let $K$ be the knot in $\mathbb{R}P^3$ obtained as the quotient of the vertical segment $V=\{(0,0,z) \in \mathbb{R}^3 : -1 \leq z \leq 1\}$ in $B^3$. As depicted in the figure below, a normal neighborhood $N(K)$ of $K \subset \mathbb{R}P^3$ is a solid torus. We claim its complement is also a solid torus. As depicted, there is a simple closed curve in $\partial N(K)$ that bounds an embedded disk $(D,\partial D) \subset (\mathbb{R}P^3 \setminus \mathring{N}(K), \partial N(K))$. Manipulating the identification diagram (not shown, but achieved by cutting $B^3 \setminus V$ along the disk and gluing via the antipodal map on $S^2 \setminus \{(0,0,\pm 1)\}$), we can see that the complement of this disk in $\mathbb{R}P^3 \setminus N(K)$ is a 3-ball. It follows that $\mathbb{R}P^3 \setminus \mathring{N}(K)$ is a solid torus. As discussed below the figure, we can apply a half-twist to $\partial N(K)$ to identify it with a standard torus, taking $\partial D$ to a $(2,1)$-curve in $\partial(S^1 \times D^2)$. This decomposition corresponds to a homeomorphism from $\mathbb{R}P^3$ to the lens space $L(2,1)$. $\square$

Left: $\mathbb{R}P^3$ is depicted as a quotient of the closed 3-ball $B^3$ via the antipodal map $a: \partial B^3 \to \partial B^3$. The vertical line depicts the knot $K \subset \mathbb{R}P^3$. Middle: The knot's normal neighborhood $N(K)$ situated inside of $\mathbb{R}P^3$. The boundary of $N(K)$, seen here as a dotted cylinder, is a torus. We can visualize a deformation retract of $\mathbb{R}P^3 \setminus N(K)$ to the equatorial $\mathbb{R}P^1=S^1$ by taking successively "larger" neighborhoods $N(K)$. Right: We see a simple closed curve in $\partial N(K)$ which bounds a disk in $\mathbb{R}P^3 \setminus K$. The antipodal map sends the "upper circle" to the "lower circle" with a rotation by $\pi$, so the depicted curve becomes a $(2,1)$-curve in $\partial N(K)$.

Answer 2

The curve $\gamma= m_2 + 2 \ l_2$ bounds a Moibus strip $M$ inside the the solid torus $T_2\subset L(2,1)$ (can you see it?) and a two-disk $D$ inside $T_1 \subset L(2,1)$. The union $S= D \cup M$ is a closed non-orientable surface homeomorphic to $\mathbb{RP}^2$, and you can easily see that its complement in $L(2,1)$ is nothing but a 3-ball.

Since $L(2,1)$ is orientable while $\mathbb{RP}^2$ is not, the normal bundle of $S$ inside $L(2,1)$ is a non-trivial line bundle. Hence, a normal neighborhood of $S \subset L(2,1)$ is homeomorphic to $\mathbb{RP}^2 \tilde{\times} I$ (the interval bundle associated to the unique non-orientable line bundle over $\mathbb{RP}^2$), and we can conclude that $L(2,1)$ is obtained from $\mathbb{RP}^2 \tilde{\times} I$ by filling its boundary 2-sphere with a 3-ball.

On the other hand, you can see directly that the normal neighborhood of $\mathbb{RP}^2 \subset \mathbb{RP}^3$ is homeomorphic to $\mathbb{RP}^2 \tilde{\times} I$ and that its complement is homeomorphic to a 3-ball. Namely, think $\mathbb{RP}^3$ as the unit ball $B^3(0,1)/\sim$ with the antipodal relation on its boundary, and decompose it as the union of the ball $B^3(0,1/2) \subset B^3(0,1)/\sim$ and its complement (homeomorphic to a non-trivial interval bundle over $\mathbb{RP}^2 = \partial B^3(0,1)/\sim$).

Hence, both $L(2,1)$ and $\mathbb{RP}^3$ are obtained from $\mathbb{RP}^2 \tilde{\times} I$ by filling its boundary 2-sphere with a 3-ball. Now, in order to conclude, we only need an important basic fact from differential topology: a 2-sphere boundary component of 3-manifold can be capped uniquely with a 3-ball.

Answer 3

If the definition of lens space $L(p,q)$ is given as two solid tori glued together, then probably you know that $S^3$ is the result of two solid tori glued together as well (meridian to longitude, and longitude to meridian), denote them $T_1$ and $T_2$.

Say $T_1$ is centered at the origin of $\mathbb{R}^3$ (and think $S^3$ as the compactification of $\mathbb{R}^3$), and consider $\mathbb{R}P^3$ as $S^3$ with antipodal points identified (~). We can see that $T_1/\sim$ is still a solid torus (there is a twist in the second step):

Consider a longitude on $T_1$, and we can see as below that after taking the quotient it becomes a curve that goes once around the meridian and twice around the longitude, let's call this curve $\alpha$:

Now think about $T_2$ and do the similar thing to a meridian curve (and the result is still a meridian).

Remember $S^3$ is the result by gluing a meridian of $T_2$ to a longitude of $T_1$, thus $\mathbb{R}P^3$ is the result by gluing meridian of a solid torus to the curve $\alpha$ on the other solid torus, but that is exactly how your $L(2,1)$ is defined.

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I personally found this an easier way to understand why they are the same with the 'gluing' definition, because this way you naturally have two solid torus from $\mathbb{R}P^3$ without building/finding them.